$\frac{8 n^{-2}}{1 n^5}=\frac{8 n}{1 n}$
The objective is to find the value of \(n\) that satisfies the equation:
\[\frac{8n^{-2}}{n^5} = \frac{8n}{n}\]
We must note that for the expressions to be defined, \(n\) cannot be \(0\), as it appears in the denominator. We will simplify both sides of the equation using exponent rules, specifically \(\frac{x^a}{x^b} = x^{a-b}\) and \(x^{-a} = \frac{1}{x^a}\).
First, we simplify the left-hand side (LHS) of the equation:
\[\frac{8n^{-2}}{n^5} = 8n^{-2-5} = 8n^{-7} = \frac{8}{n^7}\]
Next, we simplify the right-hand side (RHS) of the equation:
\[\frac{8n}{n} = 8\]
This simplification is valid for all \(n
eq 0\).
Equating the simplified LHS and RHS gives:
\[\frac{8}{n^7} = 8\]
Dividing both sides by \(8\), we get:
\[\frac{1}{n^7} = 1\]
Taking the reciprocal of both sides leads to:
\[n^7 = 1\]
For real numbers, the only value of \(n\) such that \(n^7 = 1\) is \(n = 1\). We verify that this satisfies the initial condition \(n
eq 0\). Substituting \(n=1\) back into the original equation:
LHS: \(\frac{8(1)^{-2}}{1^5} = \frac{8}{1} = 8\)
RHS: \(\frac{8(1)}{1} = 8\)
Since LHS = RHS, the solution is correct.
In a scaling scenario, consider a model where the power of a variable \(n\) represents the amplification of a signal. If you find that the amplification \(\frac{8n^{-2}}{n^5}\) needs to match a steady baseline of 8, solving for \(n\) helps determine the specific scaling factor required for your system to function correctly. This is fundamental in fields like electrical engineering and physics when normalizing data or adjusting device parameters to reach a target output level.